Question

A body of mass $10kg$ lies on a rough inclined plane of inclination $\theta ={\sin }^{-1}\left(\frac{3}{5}\right)$ with the horizontal. When the force of $30N$ is applied on the block parallel to an upward the plane, the total force by the plane on the block is nearly along:

• A. OA
• B. OB
• C. OC
• D. OD

Answer 1

Answer: (D)

OD

In this question there are three forces acting on the body as shown in the figure (1). Weight of the body mg (2). 30N force (3). normal reaction (N) along OB

Now break the force $mg$ into two mutually perpendicular directions along and perpendicular to the inclined plane.

So, as we can see that body is in equilibrium along the line $OB$

Therefore $mg\cos \theta =N$

And $mg\sin \theta$ will act along $OD$ equal to $100\times 3/5=60N$

So, net force equal to $30N$ will act along $OD$